# Proving that $\Big(\sum_{k=1}^\infty a_k\Big)^p\leq c \sum_{k=1}^\infty \beta^ka_k^p.$

Let $$p>0$$ and $$\beta>1$$ then there exists a constant $$c=c(p,\beta)$$ such for every sequence $$(a_k)_k$$ with $$a_k\geq0$$ we have

$$\Big(\sum_{k=1}^\infty a_k\Big)^p\leq c \sum_{k=1}^\infty \beta^ka_k^p$$

The case $$p\leq 1$$ is obvious here since we have $$(a_1+a_2)^p\leq a_1^p+a_2^p$$ and by induction one easily arrives at $$\Big(\sum_{k=1}^\infty a_k\Big)^p\leq \sum_{k=1}^\infty a_k^p\leq \sum_{k=1}^\infty \beta^ka_k^p$$

How to prove the case $$p>1$$?

• The inequality is invalid for the case that $p\geq1$ and $\beta=1$. For example, let $p=2$, $\beta=1$ and $a=(1,\frac{1}{2},\frac{1}{3},\ldots)$. Clearly, $\sum_{k=1}^{\infty}a_{k}=\infty$. However, $\sum_{k=1}^{\infty}\beta^{k}a_{k}^{p}=\sum_{k=1}^{\infty}\frac{1}{k^{2}}<\infty$. Oct 6 '21 at 19:55
• You are right sorry we should take $\beta>1$ Oct 7 '21 at 7:25

I consider the case that $$p>1$$ and $$\beta>1$$ and the rest is left to you. We need some preparation.

Let $$\gamma\in(0,1)$$. Let $$p>1$$. Let $$q\in(1,\infty)$$ be such that $$\frac{1}{p}+\frac{1}{q}=1$$. Define $$T:l^{p}\rightarrow l^{1}$$ by $$Ta=(a_{1}\gamma,a_{2}\gamma^{2},\ldots),$$ where $$a=(a_{1},a_{2},\ldots)\in l^{p}$$.

Firstly, we show that $$T$$ is well-defined. By Holder inequality, $$\begin{eqnarray*} \sum_{k=1}^{\infty}|a_{k}\gamma^{k}| & \leq & \left\{ \sum_{k=1}^{\infty}|a_{k}|^{p}\right\} ^{\frac{1}{p}}\left\{ \sum_{k=1}^{\infty}\gamma^{kq}\right\} ^{\frac{1}{q}}\\ & = & M||a||_{p}, \end{eqnarray*}$$ where $$M=\left\{ \sum_{k=1}^{\infty}\gamma^{kq}\right\} ^{\frac{1}{q}}<\infty$$. Note that $$M$$ depends on $$\gamma$$ and $$p$$ only. It is easy to show that $$T$$ is linear and $$||T||\leq M$$.

Now, we go back to your question. Let $$p>1$$ and $$\beta>1$$ be given. We define $$\beta_{0}=\beta^{\frac{1}{p}}>1$$, $$\gamma=\frac{1}{\beta_{0}}$$, and $$M=\left\{ \sum_{k=1}^{\infty}\gamma^{kq}\right\} ^{\frac{1}{q}}<\infty$$.

Let $$a=(a_{n})$$ be given with $$a_{n}\geq0$$. For each $$n\in\mathbb{N}$$, define $$b^{(n)}=(a_{1}\beta_{0},a_{2}\beta_{0}^{2},\ldots,a_{n}\beta_{0}^{n},0,0,\ldots).$$ Clearly $$b^{(n)}\in l^{p}$$. It follows that $$||Tb^{(n)}||_{1}\leq M||b^{(n)}||_{p}$$. That is, $$\begin{eqnarray*} \sum_{k=1}^{n}a_{k} & \leq & M\left\{ \sum_{k=1}^{n}a_{k}^{p}\beta_{0}^{kp}\right\} ^{\frac{1}{p}}\\ & = & M\left\{ \sum_{k=1}^{n}a_{k}^{p}\beta^{k}\right\} ^{\frac{1}{p}}\\ & \leq & M\left\{ \sum_{k=1}^{\infty}a_{k}^{p}\beta^{k}\right\} ^{\frac{1}{p}}. \end{eqnarray*}$$ Letting $$n\rightarrow\infty$$ yields $$\sum_{k=1}^{\infty}a_{k}\leq M\left\{ \sum_{k=1}^{\infty}a_{k}^{p}\beta^{k}\right\} ^{\frac{1}{p}}$$.

Let $$p>1$$, and let $$q=\frac{p}{p-1}$$ be the Lebesgue conjugate exponent of $$p$$, and let $$\beta>1$$. First write $$\sum_{k=1}^{\infty} a_k= \sum_{k=1}^{\infty}\frac{1}{\beta^{\frac{k}{p}}} \beta^{\frac{k}{p}} a_k.$$

Then apply Holder's inequality to get

$$\sum_{k=1}^{\infty} a_k\leq \left(\sum_{k=1}^{\infty}\left(\frac{1}{\beta^\frac{q}{p}}\right)^k\right)^{\frac{1}{q}} \left(\sum_{k=1}^{\infty} \beta^{k} a_k^p\right)^{\frac{1}{p}}.$$ The sum $$\sum_{k=1}^{\infty}\left(\frac{1}{\beta^\frac{q}{p}}\right)^k$$ converges since $$\beta^\frac{q}{p}>1$$.